CSE 21 - Winter 2014
Assignment 2
Assignment 2 Solutions
Prepared by Radheshyam Balasundaram
1 Prove that
k −1
bk − 1
∑ abi = a b − 1
i =0
Let us prove this by induction on k
Base Case
For k = 1, the term on the left hand side (LHS) has only one term, ab0 = a.
1
And, RHS evaluates to a bb−
−1 = a.
Induction Hypothesis
Assume that the statement is valid for some k, ie,
k −1
bk − 1
∑ ab = a b − 1
i =0
i
Induction Step
Now, let us prove the statement for k + 1.
k
k −1
i =0
i =0
bk
∑ abi =
∑ abi + abk
−1
+ abk (from induction hypothesis)
b−1
b k − 1 + b k +1 − b k
=a
b−1
k
+
1
b
−1
=a
b−1
=a
This proves the statement for k + 1.
1
2 Prove that
k
∑ (a + ib) =
i =0
2a + kb
( k + 1)
2
Let us prove this by induction on k
Base Case
For k = 0, LHS has only one term, a. And RHS is
2a+0
2 (0 + 1)
= a.
Induction Hypothesis
Assume that the statement is valid for some k, ie,
k
∑ (a + ib) =
i =0
2a + kb
( k + 1)
2
Induction Step
Now, let us prove the statement for k + 1.
k +1
∑ (a + ib) =
i =0
=
=
=
=
=
k
∑ (a + ib) + [a + (k + 1)b)]
i =0
2a + kb
(k + 1) + [ a + (k + 1)b] (from induction hypothesis)
2
2a(k + 1) + kb(k + 1) + [2a + (k + 1)2b]
2
2a(k + 1) + 2a + kb(k + 1) + (k + 1)2b
2
2a(k + 2) + b(k + 1)(k + 2)
2
2a + (k + 1)b
( k + 2)
2
3 Let n > 1 be an integer. In a football league there are n teams. Every
two teams have played against each other exactly once, and in match no
draw is allowed. Prove that it is possible to number the teams in such a
way that team i beats (i + 1) for i = 1, 2, . . . , n − 1
2
Use induction on n.
Base Case
For n = 2, there are only two teams. Whoever won the match between the
two will be numbered 1 and the other team is numbered 2.
Induction Hypothesis
Assume that the statement is true for n. That is, if there are n teams, we
can assign numbers 1 to n so that team i has won against team i + 1.
Induction Step
Now, let us prove the statement for n + 1. Take a tournament of n + 1
teams. Out of the n + 1 teams, consider a subset of n teams first. By induction hypothesis, we can number these n teams 1 to n such that team i + 1
has lost to team i. Now we shall ”fit in” the remaining n + 1th team in this
order.
If this n + 1th team has lost to all n teams, then assign a number n + 1.
Since it has lost to team n, the order is preserved.
Otherwise, look at the first team in the order, say team with number
j(≤ n), it has won against. Assign number j to this n + 1th team and
increment the number of all the teams that are on or after j in the order by
1.
4 Prove that 2002n+2 + 20032n+1 is divisible by 4005.
The statement is incorrect! Here’s a quick way to check the same. If a
number is a multiple of 4005, then it should also be a multiple of 5 (as 5
divides 4005). So, is 2002n+2 + 20032n+1 a multiple of 5? Let us check the
remainder of individual terms when divided by 5.
2002n+2 can be written as (2000 + 2)n+2 . As 5 divides 2000, remainder
of (2000 + 2)n+2 is same as remainder of 2n+2 when divided by 5. Similarly, 20032n+1 is equivalent to 32n+1 for remainder when divided by 5.
Now the sum boils down to 2n+2 + 32n+1 = 4 · 2n + 3 · 9n . Clearly, when
n = 2, this sum is 4 · 22 + 3 · 92 = 16 + 243 = 259 which is not divisible by
5.
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5 The Fibonacci sequence is dened as x0 = 0, x1 = 1 and xn+2 = xn+1 +
xn for all non-negative integers n. Prove that
(a) xm = xr+1 xm−r + xr xm−r−1 for all integers m ≥ 1 and 0 ≤ r ≤
m − 1;
(b) xd divides xkd for all positive integers d and k.
Part (a)
Let us prove this by induction on r in the range 0 ≤ r ≤ m − 1.
Base Case
For r = 0, xr+1 xm−r + xr xm−r−1 = x1 xm + x0 xm−1 = 1xm + 0xm−1 = xm .
Induction Hypothesis
Now, assume for some r in range 0 to m − 2 that xm = xr+1 xm−r + xr xm−r−1 .
Induction Step
Take r + 1 with r + 1 ≤ m − 1. Now, with r + 1,
RHS =
=
=
=
=
=
x r +2 x m −r −1 + x r +1 x m −r −2
( x r +1 + x r ) x m −r −1 + x r +1 x m −r −2
x r +1 x m −r −1 + x r x m −r −1 + x r +1 x m −r −2
x r +1 ( x m −r −1 + x m −r −2 ) + x r x m −r −1
x r +1 x m −r + x r x m −r −1
xm (from induction hypothesis)
Note that here r + 1 ≤ m − 1 is important because otherwise xm−r−2
will not be valid.
Part (b)
Use induction on k.
Base Case
For k = 1, xkd = xd and xd trivially divides xkd .
Induction Hypothesis
Assume for some k, xd divides xkd . That is, xkd = q · xd for some integer q.
Induction Step
Now, consider x(k+1)d . By using m = (k + 1)d and r = d in the identity in
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part (a), we get:
x ( k +1) d = x d +1 x ( k +1) d − d + x d x ( k +1) d − d −1
= xd+1 xkd + xd xkd−1
= q · xd+1 xd + xd xkd−1
= xd (qxd+1 + xkd−1 )
6 (Hard Problem) For natural numbers p and q, the Ramsey number
R( p, q) is dened as the smallest integer n so that among any n people,
there exist p of them who know each other, or there exist q of them who
dont know each other. Note that R( p, 1) = R(1, q) = 1. Prove that:
(a) R( p + 1, q + 1) ≤ R( p, q + 1) + R( p + 1, q)
p + q −2
(b) R( p, q) ≤ C p−1
In both the parts, we need induction on both p and q. A rigorous mathematical proof would involve taking an arbitrary, but fixed value for q (say,
q = q0 ) and then inducting over p. Once we have proved it for all p and
fixed q, we can use induction over q. Here, we give a sketch of proof of
induction step, which is the most critical part.
Part (a)
As induction hypothesis, assume the following two statements:
• R( p, q + 1): In a group of R( p, q + 1) people there are either p people
who know each other or q + 1 people who don’t know each other.
• R( p + 1, q): In a group of R( p + 1, q) people there are either p + 1
people who know each other or q people who don’t know each other.
Now, let us prove the following statement: In a group of R( p, q + 1) +
R( p + 1, q), there are either p + 1 people who know each other or q + 1
people who don’t know each other.
Take any group G of R( p, q + 1) + R( p + 1, q) people and select a random person x. Let F ⊆ G be the set of people x knows and E ⊆ G be the
set of people x doesn’t know.
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Now, either | F | ≥ R( p, q + 1) or | E| ≥ R( p + 1, q). Otherwise, if
| F | ≤ R( p, q + 1) − 1 and | E| ≤ R( p + 1, q) − 1, then | G | = | F | + | E| + 1 ≤
R( p, q + 1) + R( p + 1, q) − 1, a contradiction.
Case 1: | F | ≥ R( p, q + 1)
By induction hypothesis, F contains either q + 1 people who don’t know
each other, or p people who know each other. If F contains q + 1 people
who don’t know each other, then G also contains the sameset of people
and we are done. Otherwise, F contains p people who know each other.
And, x knows everyone from F and especially these p people. So, along
with x, we have found a subset of p + 1 people who know each other.
Case 2: | E| ≥ R( p + 1, 1)
The argument here is symmetric to Case 1. By induction hypothesis, E
contains either p + 1 people who know each other, or q people who don’t
know each other. If E contains p + 1 people who know each other, we are
done. Otherwise, E contains q people who don’t know each other. And, x
doesn’t know anyone from E and especially these q people. So, along with
x, we have found a subset of q + 1 people who don’t know each other.
This concludes the proof of part(a). Note that the problem can be visualized a graph problem where people represent vertices and if two people
know each other we add an edge between corresponding vertices.
Part (b)
Again, we use induction on both p and q. By induction hypothesis, assume
that:
p + q −3
1. R( p − 1, q) ≤ C p−2
p + q −3
2. R( p, q − 1) ≤ C p−1
So, we have:
R( p, q) ≤ R( p, q − 1) + R( p − 1, q) (From part (a))
p + q −3
≤ C p −1
p + q −3
+ C p −2
(From induction hypothesis)
p + q −2
= C p −1
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